phy-electromagnetism(2)

a. Current, I = V/R = 1.5 / 10 = 0.15 A


b. Current through the rod, I = 0.15 A

By F = IBL

Forrce on PQ, F = (0.15)(0.40)(0.5) = 0.03 N


c. The current flows from P to Q through the rod. When the rod starts to move, there is a change of magnetic flux of it. By Faraday's law of electromagnetic induction, there is a back e.m.f. induced in the rod which direction is opposite to that of the e.m.f. of the battery. A back current is induced since the rod has a finite resistance. When the rod moves faster, the back e.m.f. and hence the back current is larger. So, the retarding force is larger while the magnetic force remains unchanged.

When the two forces equal, they cancel out each other and hence there is no net force on the rod. And the speed of PQ attains a maximum and will not increase indefinitely.

When the back e.m.f. equals e.m.f. of battery, the terminal speed is attained.

By Faraday's law of electromagnetic induction,

V = -d(BA)/dt

V = -Bl dx/dt = -Blv

-1.5 = -(0.40)(0.50)v

Terminal speed, v = 7.5 m/s


d. Current = 0 since there is no potential difference in the rod.

(a) Current = 1.5/10 A = 0.15 A

(b) Net force = 0.4 x 0.5 x 0.15 N = 0.03 N

(c) There is a counter-force produced on PQ because of the induced current.

Hence, max speed = 1.5/(0.4 x 0.5) m/s = 7.5 m/s

(d) Zero.